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What mass of nitrogen is needed to fill an 855 L tank at STP?

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Answer:

It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.

Step-by-step explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C or 273.15 °K are used and are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

So, in this case:

  • P= 1 atm
  • V= 855 L
  • n= ?
  • R= 0.082
    (atm*L)/(mol*K)
  • T= 273.15 K

Replacing:

1 atm* 855 L= n* 0.082
(atm*L)/(mol*K) * 273.15 K

Solving:


n=(1 atm* 855 L)/(0.082(atm*L)/(mol*K) *273.15 K )

n= 38.17 moles

Being the molar mass of nitrogen N2 equal to 28 g / mol, you can apply the following rule of three: if there are 28 grams in 1 mole, how much mass is there in 38.17 moles?


mass=(38.17 moles*28 grams)/(1 mole)

mass= 1,068.76 grams

It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.

User Csibi Norbert
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